Question: Solve for $x$ : $7\sqrt{x} - 1 = 4\sqrt{x} + 6$
Explanation: Subtract $4\sqrt{x}$ from both sides: $(7\sqrt{x} - 1) - 4\sqrt{x} = (4\sqrt{x} + 6) - 4\sqrt{x}$ $3\sqrt{x} - 1 = 6$ Add $1$ to both sides: $(3\sqrt{x} - 1) + 1 = 6 + 1$ $3\sqrt{x} = 7$ Divide both sides by $3$ $\frac{3\sqrt{x}}{3} = \frac{7}{3}$ Simplify. $\sqrt{x} = \dfrac{7}{3}$ Square both sides. $\sqrt{x} \cdot \sqrt{x} = \dfrac{7}{3} \cdot \dfrac{7}{3}$ $x = \dfrac{49}{9}$